Solve for $x$ : $9\sqrt{x} - 1 = 3\sqrt{x} + 3$
Explanation: Subtract $3\sqrt{x}$ from both sides: $(9\sqrt{x} - 1) - 3\sqrt{x} = (3\sqrt{x} + 3) - 3\sqrt{x}$ $6\sqrt{x} - 1 = 3$ Add $1$ to both sides: $(6\sqrt{x} - 1) + 1 = 3 + 1$ $6\sqrt{x} = 4$ Divide both sides by $6$ $\frac{6\sqrt{x}}{6} = \frac{4}{6}$ Simplify. $\sqrt{x} = \dfrac{2}{3}$ Square both sides. $\sqrt{x} \cdot \sqrt{x} = \dfrac{2}{3} \cdot \dfrac{2}{3}$ $x = \dfrac{4}{9}$